What BEST describes these three molecules?

There are 3 naturally occurring isotopes of hydrogen and 2

DedPeter [7]

Answer:

True

Explanation:

Hydrogen has three naturally occuring isotopes and copper has two.

3 0

3 years ago

Hello,can you guys please write me an explanation on the question below, the best answer will get brainliest

Nataliya [291]

Answer:

I just umm

Explanation:

6 0

3 years ago

Arrange the following aqueous solutions in order of increasing boiling point temperature (lowest to highest temperature, with 1

Elanso [62]

Answer:

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

Explanation:

Step 1: Data given

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = Shows how much the boiling point increases

⇒i = the van't Hoff factor: Says in how many particles the compound will dissociate

⇒ Since all are aqueous solutions Kb for all solutions is the same (0.512 °C/m)

⇒m = the molality

Step 2:

0.20 m glucose

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for glucose = 1

⇒ Kb = 0.512 °C/m

⇒m = 0.20 m

ΔT = 1*0.512 * 0.20

ΔT = 0.1024 °C

0.30 m BaCl2

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for BaCl2 = Ba^2+ + 2Cl- : i = 3

⇒ Kb = 0.512 °C/m

⇒m = 0.30 m

ΔT = 3*0.512 * 0.30

ΔT = 0.4608 °C

0.40 m NaCl

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for NaCl = Na+ + Cl- : i = 2

⇒ Kb = 0.512 °C/m

⇒m = 0.40 m

ΔT = 2*0.512 * 0.40

ΔT = 0.4096 °C

0.50 m Na2SO4.

ΔT = i*Kb*m

⇒ΔT = the boiling point elevation = TO BE DETERMINED

⇒i = the van't Hoff factor for Na2SO4 = 2Na+ + SO4^2- : i =3

⇒ Kb = 0.512 °C/m

⇒m = 0.50 m

ΔT = 3*0.512 * 0.50

ΔT = 0.768 °C

0.20 m glucose < 0.40 m NaCl < 0.30 m BaCl2 < 0.50 m Na2SO4.

8 0

3 years ago

A 4.00 (±0.01) mL Class A transfer pipet is used to transfer 4.00 mL of a 0.302 (±0.004) M Cu2+ stock solution to a 100.00 (±0.0

cricket20 [7]

Answer:

concentration of diluted solution = 0.0125 ( ± 0.0002)M

Uncertainty = ± 0.0002

Explanation:

Given that

Initial volume of Cu2+ = 4.00 (±0.01) mL

Initial molarity 0f Cu2+ = 0.302 (±0.004) M

transferred to 100.00 (±0.08) Class A volumetric flask

first we get amount of water added

100.00 (±0.08) - 4.00 (±0.01) = 96 ± (0.09)

Now according to law of dilution

The concentration of Cu2+ after adding water

M1V1 = M2V2

we substitute

0.302 (±0.004) * 4.00 (±0.01) = x * 96 ± (0.09)

Now the multiplication of two digits with uncertainty is

(0.004/0.302) * 100 = 1.32% ; (0.01/4.00) * 100 = 0.25%

= [0.302 ( ± 1.32% )] * [ 4.00 ± (0.25%)]

= 1.208 ±(1.57%)

1.57/100 * 1.208 = 0.0189

so

= (1.208 ± 0.0189)

now substitute in our previous equation

1.208 ± (0.0189) = x * 96 ± (0.09)

x = 1.208 ± (0.0189) / 96 ± (0.09)

{ 0.09/96 * 100 = 0.094% }

so x = 1.208 ± (1.57%) / 96 ± (0.094% )

x = 0.0125 ± ( 1.664)

now( 1.664/100 * 0.0125)

= ± 0.000208

Hence

concentration of diluted solution = 0.0125 ( ± 0.0002)M

Uncertainty = ± 0.0002

5 0

2 years ago

The law of conservation of energy states that if the system gains heat the surroundings ________?

Arlecino [84]

The surroundings would lose heat.

8 0

2 years ago